From a point outside a circular garden, two tangents of length 15 m each are drawn. If the radius of the garden is 9 m, find the distance from the point to the center.
17.5 m
A tangent to a circle is always perpendicular to the radius drawn to the point of contact, so at the point of contact the radius and tangent meet at \(90\degree\).
Join the external point P to the centre O and to the point of contact T. Then triangle OTP is right-angled at T, with radius OT and tangent PT as the two legs and OP as the hypotenuse.
By the Pythagoras theorem, \(OP = \sqrt{PT^2 + OT^2}\).
Substitute the tangent length \(PT = 15\ \text{m}\) and radius \(OT = 9\ \text{m}\): \(OP = \sqrt{15^2 + 9^2} = \sqrt{225 + 81} = \sqrt{306}\).
Evaluating, \(\sqrt{306} \approx 17.49 \approx 17.5\ \text{m}\).
Key concept: the radius-tangent right angle turns any tangent-length problem into a right triangle solved by Pythagoras.
Hence the distance from the point to the centre is 17.5 m.
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